1. **State the problem:** Find the limit of the piecewise function $$f(x) = \begin{cases} x^2 & \text{if } x \leq 2 \\ 3x - 3 & \text{if } x > 2 \end{cases}$$ as $x$ approaches 2. 2. **Recall the limit definition for piecewise functions:** The limit at $x=2$ exists if and only if the left-hand limit and right-hand limit at 2 are equal. 3. **Calculate the left-hand limit ($x \to 2^-$):** Since $x \leq 2$, use $f(x) = x^2$. $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4$$ 4. **Calculate the right-hand limit ($x \to 2^+$):** Since $x > 2$, use $f(x) = 3x - 3$. $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 3) = 3(2) - 3 = 6 - 3 = 3$$ 5. **Compare the limits:** Left-hand limit is 4, right-hand limit is 3. Since these are not equal, the limit of $f(x)$ as $x$ approaches 2 does not exist. 6. **Summary:** - Left limit: 4 - Right limit: 3 - Limit at $x=2$: Does not exist because the two one-sided limits differ.